2nd Putnam 1939

Problem A2

Let C be the curve y = x3 (where x takes all real values). The tangent at A meets the curve again at B. Prove that the gradient at B is 4 times the gradient at A.



Trivial. [Take the point as (a,a3). Write down the equation of the tangent. Write down its point of intersection with the curve: (x3 - a3) = 3a2(x - a). We know this has a repeated root x = a. The sum of the roots is zero, so the third root is x = - 2a. Finally, 3(-2a)2 = 4 times 3 a2.]



2nd Putnam 1939

© John Scholes
4 Sep 1999